Question

Prove the given expression, \[\cos [ta{{n}^{-1}}\{sin(co{{t}^{-1}}x)\}]=\sqrt{\dfrac{1+{{x}^{2}}}{2+{{x}^{2}}}}\] .

Answer 1

Hint: In this question, we have multiple trigonometric functions. So we have to do trigonometric conversion multiple times. We have inverse tan function and inverse cot function. Assume, \[\theta ={{\cot }^{-1}}x\] and then transform \[\cot \theta \] into \[\sin \theta \] . Also assume \[\beta ={{\tan }^{-1}}\left( \dfrac{1}{\sqrt{1+{{x}^{2}}}} \right)\] and then transform \[\tan \beta \] into \[\cos \beta \] . Now, it can be solved easily.

Complete step-by-step solution -
Solve this question, step by step.
Treat \[\sin (co{{t}^{-1}}x)\] as the first part and then simplify this.
So, first of all, we have to solve \[\sin (co{{t}^{-1}}x)\].
Let us assume,
\[\theta ={{\cot }^{-1}}x\]
\[\Rightarrow \cot \theta =x\]…………..(1)
We have, \[\cot \theta =\dfrac{\text{base}}{\text{Height}} \],

Base = x,
Height= 1,
Using Pythagoras theorem Hypotenuse =\[\sqrt{{{\left(\text{height} \right)}^{2}}+{{\left(\text{Base} \right)}^{2}}}\], we get
Hypotenuse= \[\sqrt{1+{{x}^{2}}}\]
\[\sin \theta =\dfrac{\text{height}}{\text{hypotenuse}}\]
\[\sin \theta =\dfrac{1}{\sqrt{1+{{x}^{2}}}}\]
\[\Rightarrow \theta ={{\sin }^{-1}}\left( \dfrac{1}{\sqrt{1+{{x}^{2}}}} \right)\]………………(2)
According to the question, we have \[\cos [ta{{n}^{-1}}{\sin({\cot}^{-1}}x)\}]\]………….(3)
From equation(1), we have \[\theta ={{\cot }^{-1}}x\] .
Substituting equation(1) in equation(3), we get \[\cos [ta{{n}^{-1}}\{sin\theta \}]\]…………..(4)
Now, using equation(2), equation(4) can be written as
\[\begin{align}
  & \sin ({{\sin }^{-1}}\left( \dfrac{1}{\sqrt{1+{{x}^{2}}}} \right)) \\
 & =\dfrac{1}{\sqrt{1+{{x}^{2}}}} \\
\end{align}\]
Our equation may be written as,
\[\begin{align}
  & \cos \left[ {{\tan }^{-1}}\left\{ \sin \left( {{\sin }^{-1}}\left( \dfrac{1}{\sqrt{1+{{x}^{2}}}} \right) \right) \right\} \right] \\
 & =\cos \left[ {{\tan }^{-1}}\left( \dfrac{1}{\sqrt{1+{{x}^{2}}}} \right) \right] \\
\end{align}\]
We have simplified our equation given in the question.
Now, we have to solve the equation, \[\cos \left[ {{\tan }^{-1}}\left( \dfrac{1}{\sqrt{1+{{x}^{2}}}} \right) \right]\]……….(5)
For this, we have to convert the inverse tan function into inverse cosine function.
Similarly, let us assume,
\[\beta ={{\tan }^{-1}}\left( \dfrac{1}{\sqrt{1+{{x}^{2}}}} \right)\]……………..(6)
Solving equation(6), we get \[\tan \beta =\dfrac{1}{\sqrt{1+{{x}^{2}}}}\]…………..(7)
We also know the identity, \[{{\sec }^{2}}\beta -{{\tan }^{2}}\beta =1\] .
\[\sec \beta =\sqrt{1+{{\tan }^{2}}\beta }\]……………..(8)
Using equation(7) and substituting it in equation(8), we get
\[\begin{align}
  & \sec \beta =\sqrt{1+{{\tan }^{2}}\beta } \\
 & =\sqrt{1+\dfrac{1}{1+{{x}^{2}}}} \\
 & =\sqrt{\dfrac{1+{{x}^{2}}+1}{1+{{x}^{2}}}} \\
 & =\sqrt{\dfrac{2+{{x}^{2}}}{1+{{x}^{2}}}} \\
\end{align}\]
We also know that, \[\cos \beta =\dfrac{1}{\sec \beta }\] .
Using this formula we can find \[\cos \beta\] .
\[\cos \beta =\dfrac{\sqrt{1+{{x}^{2}}}}{\sqrt{2+{{x}^{2}}}}\]…………….(9)
Substituting equation(6) in equation(5), we get
\[\cos (ta{{n}^{-1}}\left( \dfrac{1}{\sqrt{1+{{x}^{2}}}} \right))\]
\[=\cos \beta \]
Substituting the value from equation(9), we get
\[\cos \beta =\sqrt{\dfrac{1+{{x}^{2}}}{2+{{x}^{2}}}}\]
Therefore,LHS=RHS
Hence, proved.

Note: In this question, we have to transform one trigonometric function into other trigonometric functions multiple times. So, one can easily make a mistake in the calculations involved in the transformation. It will be easier if we transform the functions using the right-angled triangle and Pythagoras theorem.
  

Pythagoras theorem,
Hypotenuse =\[\sqrt{{{\left( \text{height} \right)}^{2}}+{{\left(\text{Base}\right)}^{2}}}\] .
Using this formula, we can find height, base and hypotenuse. Now, transformation can be done easily.